- #1
LusTRouZ
- 3
- 0
I'm not understanding why my answer is wrong
[tex]\int[/tex]tan^3(x) dx
This is the solution I've been getting for this problem, but I notice you get a different answer when you let u = tan(x) and du = sec^2(x) dx
tan^2(x) = (sec^2(x) - 1)
[tex]\int[/tex]tan(x) (sec^2(x) - 1) dx
[tex]\int[/tex]tan(x) sec^2(x) - [tex]\int[/tex]tan(x) dx
. // u = sec(x) du = sec(x)tan(x) dx
[tex]\int[/tex]u du - [tex]\int[/tex]sin(x) dx[tex]/[/tex]cos(x)
. // z = cos(x) zu = -sin(x) dx
[tex]\int[/tex]u du - [tex]\int[/tex]-zu[tex]/[/tex]z
(1[tex]/[/tex]2)u^2 - (-ln |z|)
(1[tex]/[/tex]2)sec^2(x) + ln |cos(x)|
Homework Statement
[tex]\int[/tex]tan^3(x) dx
This is the solution I've been getting for this problem, but I notice you get a different answer when you let u = tan(x) and du = sec^2(x) dx
Homework Equations
tan^2(x) = (sec^2(x) - 1)
The Attempt at a Solution
[tex]\int[/tex]tan(x) (sec^2(x) - 1) dx
[tex]\int[/tex]tan(x) sec^2(x) - [tex]\int[/tex]tan(x) dx
. // u = sec(x) du = sec(x)tan(x) dx
[tex]\int[/tex]u du - [tex]\int[/tex]sin(x) dx[tex]/[/tex]cos(x)
. // z = cos(x) zu = -sin(x) dx
[tex]\int[/tex]u du - [tex]\int[/tex]-zu[tex]/[/tex]z
(1[tex]/[/tex]2)u^2 - (-ln |z|)
(1[tex]/[/tex]2)sec^2(x) + ln |cos(x)|
Homework Statement
Homework Equations
The Attempt at a Solution
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